RPA and The Shell Model

Transcription

1 RPA and The Shell Model Calvin W. Johnson San Diego State University Ionel Stetcu Louisiana State University & University of Arizona I. Stetcu and C. W. Johnson, Phys. Rev. C (2002) C. W. Johnson and I. Stetcu, Phys. Rev. C 66, (2002) I. Stetcu and C. W. Johnson, Phys. Rev. C. 67, (2003) I. Stetcu and C. W. Johnson, Phys Rev. C 69, (2004) Johnson, Vazquez, and Stetcu, in preparation This work was supported by grants from the Department of Energy INT November

2 Motivational Slide density functional theory 0hω shell model ab initio methods INT November

3 Even MORE Motivation: We believe there are good prospects to develop a better global theory [of nuclear energies]...treating correlation energies by RPA. * * Bertsch & Hagino nucl-th/ Phys.Atom.Nucl. 64 (2001) 588/ Yad.Fiz. 64 (2001) 646 INT November

4 Systematic Investigation of RPA or Outline of my talk! Review of RPA Overview of previous benchmarking of RPA Shell model context (in brief) and our code SHERPA Four elemental tests: (1) Correlation energies (2) Scalar observables and restoration of symmetries (3) Transitions in RPA and pnrpa and sum rules (4) Collapse of RPA Conclusions... and future work INT November

5 Review of RPA RPA models excited states as small oscillations about the mean-field. There are many ways to derive the random phase approximation (RPA): eqns of motion, time-dependent Hartree-Fock, linear response... INT November

6 The nuclear landscape Hartree-Fock based upon variational principle: minimize Ψ Ĥ Ψ Here Ψ is a Slater determinant, a product of single-particle wfns which can be written as a vector : Find minimum on nuclear energy surface = Hartree-Fock! c1 0s 1/2, m=+ ½ c2 0s c 1/2, m=- ½ 3 0p 3/2, m=+ 3/2 M c N INT November

7 The nuclear landscape There are many ways to derive the random phase approximation (RPA): eqns of motion, time-dependent Hartree-Fock, linear response... I prefer quantization of the energy surface: Ψ HF We often approximate a potential by a harmonic oscillator V ( x) V ( x x ) + 2 V ( x0)( x 0) INT November

8 The nuclear landscape To parameterize the energy surface, exponentiate the particle-hole operator: Then consider r E( Z) r r Ψ( Z) Hˆ Ψ( Z) r r Ψ( Z) Ψ( Z) exp mi Z ˆ* ˆ miamai Ψ0 = ) Ψ(Z r Thouless theorem = Hartree-Fock energy is at the minimum, Z=0 This defines an energy surface: E(Z) INT November

9 INT November The nuclear landscape ) ( ) ( ) ( ˆ ) ( ) ( Z Z Z H Z Z E r r r r r Ψ Ψ Ψ Ψ = ( ) + = * * * * 2 1 (0) Z Z Z Z E r r r r A B B A We can expand to quadratic order about Z=0: We can treat this as a harmonic oscillator by replacing numbers Z with (boson) operators b ( ) + = + + b b b b E ˆ ˆ ˆ ˆ (0) * * 2 1 A B B A

10 The nuclear landscape The final step (much math) is to write this in diagonal form: using a Bogoliubov (quasiboson) transformation and ( ˆ ) + 1 β E( 0) + 2 TrA hωλ λ λ + A B * ˆ β λ Y 1 β ˆ 2 λ B A * r X r Y λ λ = = hω mi λ X r X r Y bˆ mi, λ mi mi, λ There is a correction to the Hartree-Fock energy due to zero-point motion or, correlations among the nucleons: E RPA = + Ω 1 E( 0) TrA h 2 λ λ λ 1 () λ 2 INT November bˆ + mi The RPA matrix equation

11 Interpretation of matrix A: A mi,nj = <m i -1 H n j -1 > = <Ψ HF a i a m H a n a j Ψ HF > = matrix elements of H between 1p-1h states Solving A alone is the Tamm-Dancoff Approximation (TDA) Interpretation of matrix B: B mi,nj = <Ψ HF H m i -1 n j -1 > <Ψ HF H a m a i a n a j Ψ HF > = matrix elements of H between 2p-2h states and the HF state INT November

12 RPA as generalized harmonic oscillator From the h.o. creation/annihilation operators β λ, β λ, we can go back to generalized coordinates/momenta P λ, Q λ H RPA = λ Pˆ 2M 2 λ λ M λ Ω 2 λ Qˆ 2 λ The RPA wfn can then be written as a Gaussian...IF Ω λ > 0 Ψ RPA exp λ M Ω 2h Q λ λ 2 λ There can be zero-frequency modes...in the nuclear landscape, the HF energy does not change in those directions. These arise from broken symmetries (translation, rotation). Such zero-frequency modes must be treated with care... INT November

13 Degenerate nuclear landscapes translational rotational We will revisit this issue in a few minutes... INT November

14 Historical validation of RPA INT November

15 Benchmarking RPA Despite its widespread use, RPA has generally been only tested against toy models A typical exampe are Lipkin-type models: Parikh & Rowe, Phys Rev 175 (1968) 1293 Hagino & Bertsch, PRC C 61 (2000) acts like parity conservation 2e N particles, each either up or down... simple quasispin Hamiltonian The 2-body interaction promotes or demotes 2 particles at a time Your basic Lipkin model has only 2 independent parameters: N, the number of particles and ratio of 2-body to single-particle spitting, V/ε INT November

16 Benchmarking RPA -4 g.s. energy exact "HF" "HF+RPA" I ll return to this point later (N-1)V/e INT November

17 Benchmarking RPA Other tests of RPA... two-level pairing Hagino & Bertsch, Nucl. Phys. A679 (2000) 163 schematic interaction in small SM space Ullah & Rowe, Phys. Rev. 188 (1969) handful of others... INT November

18 The Shell-Model Context and a flexible RPA code in the shell model Diagonalization of a Hamiltonian in a shell-model basis yields exact (for that space) and nontrivial numerical results Let s compare RPA against these numerical shell-model results INT November

19 How a shell-model code works Shell-model codes, such as OXBASH, ANTOINE, or REDSTICK, writes the Schrodinger eqn as a matrix eigenvalue equations One defines a single-particle basis......selects a valence space......puts in valence nucleons......possibly assuming an inert core. Let s ask an expert... INT November

20 How a shell-model code works The basis is the set of Slater determinants of all (sort of) possible configurations in the valence space The interaction Hamiltonian is specified as single-particle energies plus two-body matrix elements (the residual interaction ) These are read in to the program as a list of numbers INT November

21 How a shell-model code works The hard part is actually computing efficiently the many-body matrix elements from the two-body matrix elements The final result is the low-lying energy spectrum and the corresponding wavefunctions (the coefficients in the Slater determinant basis) INT November

22 I have an idea! Let s write an RPA code using exactly the same shell-model input! INT November

23 SHEll-model RPA code (Stetcu PhD LSU 2003) Shell-model input compatible with REDSTICK: list of single-particle orbits (0s 1/2, 0p 3/2 etc.) list of two-body matrix elements < ab; JT H cd;jt > fair to compare output with REDSTICK results Fully self-consistent Hartree-Fock: no restrictions on Slater determinant arbitrary deformations within model space (except, wfns purely real) Standard RPA: solve matrix RPA equations see rotation of deformed HF state as zero-frequency modes; option to do pnrpa INT November

24 The First Element: RPA Correlation Energies (g.s. Binding energies beyond the mean-field) INT November

25 Results: Correlation energies Upper energies = HF energy E (MeV) Si S Ne 36 Ar Mg E (MeV) F Ne Na Si P S Cl Ar Mg Al All energies relative to exact SM diagonalization g.s. Lower energies = HF+RPA correlation energy Poorest results for single-species calculations (oxygen, calcium isotopes) INT November

26 Results: Correlation energies Comments: space # nuclides rms err (kev) sd (p+n) sd oxygen pf (p+n) pf calcium no obvious correlation of error with: magnitude of deformation (results better for deformed than spherical) goodness of HF state (overlap with exact SM g.s.) not clear that HFB+QRPA will improve the situation: if we set all pairing matrix elements = 0, results get worse. INT November

27 The Second Element: Scalar observables and restoration of broken symmetries INT November

28 In the same way as computing the ground state correlation energy, one can derive RPA corrections to the g.s. expectation value of scalar ( Hamiltonian-like ) operators Example: g.s. values of < J 2 >, < Q 2 >, < S 2 > or < r 2 > This provides a useful test of RPA... for example, a deformed HF state has < J 2 > HF 0... is < J 2 > HF+RPA = 0? Remember: we expand these operators only to 2 nd order INT November

29 Results: g.s. expectation values <J 2 > Nucleus HF RPA SM 20 Ne Mg Na Mg Na V O O Notice unphysical value! and (rarely) RPA can make expectation value worse this is because of truncating approximation at 2nd order (not fully treating Pauli principle) Conclusion: RPA is often an improvement over HF, but not a great improvement INT November

30 Hey! I m sure I read somewhere that RPA restores broken symmetries! What s going on? Shouldn t I get J 2 exact? A: RPA does restore broken symmetries but in a restricted fashion Specifically, it restores the symmetry to the energy landscape but not in the RPA wavefunction! INT November

31 Remember the energy landscape TDA does not yield Goldstone modes, The Hartree-Fock energy is independent of orientation... but RPA does...this is the restoration of the broken symmetry since TDA and RPA are expansion in all particle-hole excitations, this independence ought to be reflected as Goldstone modes (zero-energy excitations) INT November

32 What about the RPA wavefunction? The RPA wavefunction is not a symmetry-projected wavefunction... RPA is a perturbative expansion about the mean-field state... and so RPA only restores the symmetry in the vicinity of the mean-field state, not globally. (Marshalek & Weneser, Ann. Phys. (N.Y.) 53, 569 (1969)) Remember that RPA wavefunction is essentially a Gaussian (harmonic oscillator) so cannot model a global wfn for, say, linear momentum (plane wave) or angular momentum (Wigner D-function) To truly restore a broken symmetry, one needs a global or topological approach... I have some ideas! INT November

33 The Third Element: Transitions and sum rules INT November

34 RPA transition strengths exact shell model results Transition Strength / MeV Isovector E GT SF Ex. Energy (MeV) RPA results Notice the bump for each transition: a resonance INT November

35 RPA transition strengths exact shell model results Transition Strength / MeV Isovector E GT SF Ex. Energy (MeV) Transition Strength / MeV RPA results Isoscalar E Ex. Energy (MeV) missing strength INT November

36 RPA transition strengths Hey! Missing strength? What is this? Isn t there some sort of energy-weighted sum rule that should be going here? 0.10 Transition Strength / MeV 0.05 Isoscalar E Ex. Energy (MeV) missing strength INT November

37 Sum Rules and Rule-Breakers Hey! Missing strength? What is this? Isn t there some sort of energy-weighted sum rule that should be going here? Let F be a transition operator; then the energy-weighted sum rule states that [ Fˆ [ Hˆ, Fˆ ] HF = hω 1 2 HF ˆ λ 0 F λ λ RPA 2 This theorem is proven in many text-books...but is wrong! INT November

38 Sum Rules and Rule-Breakers The proof assumes no Goldstone (zero-energy) modes if one rederives it using those Goldstone modes one gets a correction [ Fˆ [ Hˆ, Fˆ ] HF = hω 1 2 HF ˆ λ 0 F λ + ν, Ω= 0 1 2M ν [ F, Pˆ ] ν 2 λ RPA Correction term (Stetcu, 2003) 2 INT November

39 Sum Rules and Rule-Breakers The within proof the intrinsic assumes state) while no Goldstone RPA models (zero-energy) transitions within modes a vibrational band if one rederives it using those Goldstone modes one gets a correction This is bolstered by the fact that we see missing strength (in even-even nuclides) 1 [ [ ] for E2 transitions but not for, 2 ˆ ˆ ˆ = 2 HF F H, F HF hω ˆ say, spin-flip ( J=1) transitions λ 0 F λ (because rotational band only allows J RPA 2 ) λ + ν, Ω= 0 1 2M ν The missing strength can be interpreted as transitions within a rotational band (that is, [ F, Pˆ ] ν 2 Correction term (Stetcu, 2003) INT November

40 Sum Rules and Rule-Breakers The missing strength can be interpreted as transitions within a rotational band (that is, The within proof the intrinsic assumes state) while no Goldstone RPA models (zero-energy) transitions within modes a vibrational band In addition, if we choose a spherical state (no Goldstone modes) rather than a deformed state, we regain the missing strength INT November

41 We also looked at Gamow-Teller transitions in pnrpa here there have been previous detailed comparisons with the shell model, but using spherical pnqrpa the central question: which is more important, pairing or deformation? INT November

42 Other s Previous Work: pn-qrpa A number of papers compared spherical pn-qrpa against exact shell model calculations of Gamow-Teller strengths Laurtizen, Nucl Phys A489 (1988) 237. Zhao & Brown, PRC 47 (1993) 2641 Running sum of GT strength INT November

43 Other s Previous Work: pn-qrpa Most likely explanation: pn-qrpa fails to sufficiently smear the Fermi surface insufficient fragmentation of GT strength QRPA 2p-2h in spherical SM Auerbach, Bertsch, Brown & Zhao, Nucl Phys A556 (1993) 190 INT November

44 Our Calculations: Deformed pn-rpa We redid this work, eschewing pairing correlations in favor of unrestricted deformations our pn-rpa pn-qrpa (Lauritzen) exact shell model INT November

45 Our Calculations: Deformed pn-rpa Clearly deformed pn-rpa is superior in total strength as well as general agreement Σ B(GT) β + β 26 Mg Σ B(GT) E (MeV) exact shell model our pn-rpa pn-qrpa (Lauritzen) E (MeV) INT November

46 Our Calculations: Deformed pn-rpa Not only deformation, but triaxiality improves the result INT November

47 Our Calculations: Deformed pn-rpa 2 8 Σ B(GT) 1 β + β 24 Na Σ B(GT) E (MeV) E (MeV) INT November

48 The Fourth Element: Collapse of RPA at phase transitions INT November

49 Collapse at phase transitions Example from the Lipkin model... spherical deformed -4 E gs g.s. energy exact "HF" "HF+RPA" (N-1)V/e collapse of RPA INT November

50 Collapse at phase transitions g.s. energy spherical E gs At the transition point several things happen: exact "HF" "HF+RPA" deformed -- At least one RPA frequency 0 -- the corresponding h-p amplitudes Y mi -- the correlation energy - while other observables go to ± (N-1)V/e collapse of RPA INT November

51 Collapse at phase transitions Do we see this with SHERPA? We induced a shape transition in 28 Si (which normally has an oblate HF state) by lowering the 0d 5/2 single-particle energy until it became spherical spherical deformed No collapse of RPA! What s going on? INT November

52 Collapse at phase transitions I know what s happening! I wrote about it in D.Thouless, Nucl. Phys. 22, 78 (1961) No collapse of RPA! What s going on? INT November

53 Collapse at phase transitions There are first-order and second-order transitions! First order: coexistence of stable solutions, no collapse Second order: no coexistence collapse!! INT November

54 Collapse at phase transitions There are first-order and second-order transitions! First order: coexistence of stable solutions, no collapse Second order: no coexistence = collapse!! Even-parity transitions (such as quadrupole) should be first order! while odd-parity transitions should be second order! INT November

55 Collapse at phase transitions Quadrupole shape transitions are first order g.s. energy (MeV) Lipkin model is 2 nd order and is more analogous to mixing of parity across major shells Example: 0p1/2-0d5/2 model space displays true collapse "16O" 16O in p1/2-d5/2 space in p1/2-d5/2 space HF exact (SM) HF+RPA -12 HF 50 HF+RPA shell separation (MeV) shell separation (MeV) < QQ > INT November

56 Collapse at phase transitions Quadrupole shape transitions are first order 28 Si in sd shell INT November

57 CONCLUSIONS We have realized RPA in a non-trivial shell model framework Tests of RPA show it to be a modest approximation to the full many-body diagonalization One should speak carefully of topics such as symmetry restoration and collapse in RPA It may be interested to continue this work onto extended RPA and other approaches such as generator-coordinate (which may yield fast, cheap solutions with ab initio Hamiltonians) INT November